Question

One cubic meter of an ideal gas is at a temperature of 10^5 n/m^2 and temperature 300 k the gas is allowed to expand at constant pressure to twice its volume by supplying heat of the change in internal energy in this process is 10^4j then the heat supplied is

Answer 1

pv=nRT

10^5*1=n(83.14)(300)

n=40

du=nCvdt(V∝T)⇒Cv=0.833

ΔQ=nCpdt=n(Cv+R)dt

dQ=11*10^4

sorry for not doing calculations

Answer 2

Answer:∆Q=?

∆U=10⁴

P= 10⁵

Explanation:

∆Q=∆U+∆W----->(1)

==> ∆W=PdV

∆Q=∆U+PdV

V1=1 , V2=2V1

Substitute in one

= 10⁴+10⁵(2-1)

= 10⁴+10⁵×1

= 10⁴(1+10)

= 10⁴(11)

==> ∆Q=11×10⁴