Question

to an ideal triatomic gas 800 cal heat energy is given at constant pressure. if vibrational mode is neglected, then energy used by gas in work done against surrounding is a) 200cal b) 300cal c) 400cal d) 60cal.

Answer 1

Energy used by gas in work done against surrounding is c) 400 cal

Explanation :

According to the first law of thermodynamics

Q = ΔU - W

Where ΔU is the change in the internal energy and W is the work done,

the work done = W = PΔV

Initially internal energy of N moles of diatomic gas is

U₁ = N6RT₁/2 = 3NRT₁

Finally after heating the internal energy of the gas will be,

U₂ = N6RT₂/2 = 3NRT₂

Thus change in internal energy,

ΔU = 3NRT₂ - 3NRT₁ = 3NR(T₂ - T₁) = 3PΔV = 3W

Hence

Q = ΔU - W

= 3W - W

=2W

=> W = Q/2 = 800/2 = 400 cal

Hence energy used by gas in work done against surrounding is c) 400 cal

Answer 2

Answer:

Explanation: option a is correct for reference see the attachment .Hope it helps you..☺